Re: [xsl] Does the count() function require access to the whole subtree?

[Michael Kay <mike@xxxxxxxxxxxx>]

> My first question in other words: Is count( x ) always streamable, no
> matter what x (or a $x or...) is?

No. Counter-example: count(following-sibling::x) is not streamable.

More specifically, count(X) is streamable (more specifically, its posture is
grounded) if the posture of X is grounded, climbing, striding, or crawling,
but not if it is roaming.

Sorry - I know that's a lot of new jargon to absorb on a Monday morning!

Michael Kay
Saxonica


>
> The "positioned" was referring to a quote from Roger, see his post, please.
>
> -W
>
>
> On 13/01/2014, Michael Kay <mike@xxxxxxxxxxxx> wrote:
>>
>> On 13 Jan 2014, at 11:59, Wolfgang Laun <wolfgang.laun@xxxxxxxxx> wrote:
>>
>>> With growing insecurity ;-)
>>>
>>> My understanding is that count($x) may be called on the construction
>>> of a sequence which (the construction) is not streamable, and that
>>> calling count(...) on it does not make it streamable.
>>>
>>> Another thing: calling count(...) doesn't require to be positioned
>>> anywhere.
>>>
>>
>> Sorry, but I'm afraid I don't even understand what you're saying/asking
>> here. What do you mean by an expression being positioned?
>>
>> What we are discussing, is in simplified terms, the fact that count(//x)
is
>> streamable, but data(//x) is not. Here //x is a "crawling" expression -
one
>> that selects nodes which may overlap each other. When an expression
returns
>> (potentially) overlapping nodes, the W3C spec says you can apply
inspection
>> operations like count() to those nodes, but you cannot apply absorption
>> expressions like data(), because doing so would require buffering.
>>
>> Michael Kay
>> Saxonica